I am trying to make a request to the Stripe API (which is a payment API) to get a json response which the API gives. I tried the following code I found in a Python course, but as the API needs to authenticate to get the response, I don't know how can I add that data in the request.

Making it in a curl request will be like this:

curl https://api.stripe.com/v1/checkout/sessions   -u pk_test_51LoUz3BgfM0E8ZiCV8UO79gw8zw7fhSgHUEAVj4wS7igs5D4kKiNsxXGeKQEUhorImNUiCxCNNtidwNkhFPUHP4i0060lsvsbw:   -d success_url="http://127.0.0.1:5500/pages/success.html"   -d cancel_url="http://127.0.0.1:5500/pages/nosuccess.html"   -d "line_items[0][price]"=price_1LvNRkBgfM0E8ZiCTSiaNvNL   -d "line_items[0][quantity]"=1 -d mode=subscription -d client_reference_id="123" -d customer_email="[email protected]" -d client_reference_id="tokenized" -d phone_number_collection["enabled"]=true

This will create a new checkout session in your stripe account and the response is a json with the info of the checkout session.

The authorisation I use in the curl request is the '-u' value of the publishable_key followed by ':' which means no password is required

But when I try to make this in Python I get the following error:

Traceback (most recent call last):
  File "/home/pau/Desktop/bsnbl/Backend/borrar.py", line 17, in <module>
    respuesta = urllib.request.urlopen(request)
  File "/usr/lib/python3.8/urllib/request.py", line 222, in urlopen
    return opener.open(url, data, timeout)
  File "/usr/lib/python3.8/urllib/request.py", line 531, in open
    response = meth(req, response)
  File "/usr/lib/python3.8/urllib/request.py", line 640, in http_response
    response = self.parent.error(
  File "/usr/lib/python3.8/urllib/request.py", line 569, in error
    return self._call_chain(*args)
  File "/usr/lib/python3.8/urllib/request.py", line 502, in _call_chain
    result = func(*args)
  File "/usr/lib/python3.8/urllib/request.py", line 649, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 401: Unauthorized

Python code:

import urllib.request
import json

# Debido a cambios en la libreria ahora se deben pasar algunos cabeceros html
paymentInfo = '-u pk_test_51LoUz3BgfM0E8ZiCV8UO79gw8zw7fhSgHUEAVj4wS7igs5D4kKiNsxXGeKQEUhorImNUiCxCNNtidwNkhFPUHP4i0060lsvsbw:   -d success_url="http://127.0.0.1:5500/pages/success.html"   -d cancel_url="http://127.0.0.1:5500/pages/nosuccess.html"   -d "line_items[0][price]"=price_1LvNRkBgfM0E8ZiCTSiaNvNL   -d "line_items[0][quantity]"=1 -d mode=subscription -d client_reference_id="123" -d customer_email="[email protected]" -d client_reference_id="tokenized" -d phone_number_collection["enabled"]=true'
res = bytes(paymentInfo,'utf-8')
print(str(type(res))) 
request = urllib.request.Request(
    'https://api.stripe.com/v1/checkout/sessions',
    data=res,
    headers={
        'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/35.0.1916.47 Safari/537.36'
    }
)
respuesta = urllib.request.urlopen(request)
print('1. ',respuesta)
cuerpo_respuesta = respuesta.read()
print(cuerpo_respuesta)
#Procesamos la respuesta json
json_respuesta = json.loads(cuerpo_respuesta.decode("utf-8"))
print(json_respuesta)

For asking the question I am using the publishable key of my Stripe account for security reasons, if you try the code with the publishable code it will ask you for the secret key, which I can't give through here. Sorry for the inconvenience

CodePudding user response：

Using the urllib module, you'll need to set Authorization header on your request to add the API key as per the Stripe documentation.

request = urllib.request.Request('https://api.stripe.com/v1/checkout/sessions',
  data=res,
  headers={
    'Authorization: Bearer sk_xxx'
  }
}

CodePudding user response：

I found the Python API in this link: https://stripe.com/docs/api/checkout/sessions/create?lang=python

By default it appears as curl but when you add lang=python you get the python API code